## Thursday, April 19, 2007

### 12.1 Introduction to Limits

12.1 Introduction to Limits

Watch the tutorials on lessons 1 - 3 on the following website.

http://www.calculus-help.com/funstuff/phobe.html

Lesson 1. What is a limit
Lesson 2. When does a limit exist.
Lesson 3. How do you evaluate a limit?

______________________________________________________

The Limit Concept - this notion of a limit is a fundamental concept of calculus.

Example 1: You are given 32 feet of fence and are asked to form a rectangular pen whose area is as large as possible. What is the largest area the pen would cover and what dimensions should the pen be?

If we let w be the width of the pen and l be the length of the pen.
Therefore we know that 2w + 2l = 32 for the perimeter.

Therefore if we solve for the length, l = 16 - w

then the area would be

A = lw
A = (16 - w)(w)
A = 16w - w2

Using this equation, we can experiment with different values of w to see how to obtain the maximum area.

(w, A) = {(1, 15), (2, 28), (3, 39), (4, 48), (5, 55), (6, 60), (7, 63), (8, 64), (9, 63), (10, 60), (11, 55), (12, 48), (13, 39) ...}

so as you can see it appears that (8, 64) is a maximum point so therefore when w = 8, the maximum area would be 64. The dimensions of the pen would be an 8 by 8 pen.

In limit terminology, you can say that "the limit of A as w approaches 8 is 64"
This is written as

lim A = 64
w®8

Definition of Limit:

If f(x) becomes arbitrarily close to a unique number L as x approaches c from either side, the limit of f(x) as x approaches c is L. This is written as

lim f(x) = L.
x®c

Example 2: Estimating a Limit Numerically

lim (4 - 3x)
x®3

let's plug in some values and see what happens when x approaches 3.
(x,f(x)) = {(2.9, -4.7), (2.99, -4.97), (2.999, -4.997), (3, ?), (3.001, -5.003), (3.01, -5.03), (3.1, -5.3)}

So we can conclude that as x approaches 3, f(x) approaches -5.

So lim (4 - 3x) = -5
x®3

this graph is continuous. For graphs that are not continuous, finding a limit can be more difficult.

Example 3: Estimating a Limit Numerically:

lim (x+1)/(x2-x-2)
x®-1

Again, let's look at some values for x around -1.

(x, f(x)) = {(-1.1, -.3226), (-1.01, -.3322), (-1.001, -.3332), (-1, error),
(-.999, -.3334), (-.99, -.3344), (-0.9, -.3448)}

So as you can see in the table when x = -1, f(x) = error because if you place -1 in for x in the equation, f(x) = 0/0 which is impossible to have because you cannot divide by 0.

But if you place values close to -1, f(x) ® -1/3 so

lim (x+1)/(x2-x-2) = -1/3
x®-1

Now try:

lim (x + 1)/(x2-x-2)
x®2

again f(2) = 3/0 which you cannot do so:

when we look at the value of f(x) at values of x around 2, you have two different answers.

1. as x approaches 2 from the left side:

lim (x + 1)/(x2-x - 2)
x®2-

you can see the limit approaches -¥

2. as x approaches 2 from the right side:

lim (x + 1)/(x2-x - 2)
x®2+

you can see the limit approaches +¥

so since f(x) is not bound as x approaches 2, you can conclude that the limit does not exist at x = 2.

Example 4:

lim (3x2-12)/(x-2)
x®2

Again plugging x = 2, f(2) = 0/0 which is impossible to have so

(x, f(x)) = {(0, 6), (1, 9), (2, error), (3, 15), (4, 18), (5, 21), ...} so you can see that it is suppose to be 12. we can do this if we use factoring:

(3x2 - 12)/(x - 2) =

(3(x2 - 4))/(x - 2) =

( 3 (x + 2)(x - 2))/ (x - 2) =

3 (x + 2) because the (x - 2) divide out so we can simplify the limit to look like:

lim 3(x + 2)
x®2

= 3 ( 2 + 2) = 12

Example 5:

lim (cos (x) - 1)/ x
x®0

again when x =0, f(0) = error but when we look at the graph, as x approaches 0 from the left side and the right side, f(x) = 0 so we can conclude

lim (cos (x) - 1)/ x = 0
x®0

Example 6:

lim (cos (1/x))
x®0

Try using different windows:
let x be between -2p and +2p and y between -1 and +2

Now zoom in alittle: let x be between -1 and +1 by .1

Now zoom in more: let x be between -.1 and +.1 by .01

Continue this, what do you notice?

The smaller the window, so the closer to x = 0 you get, the graph does not approach any particular number (it goes between +1 and -1).

Therefore this limit does not exist because no matter how close you are to zero, it is possible to choose values of x1 and x2 such that cos(1/x1) and cos(1/x2) that do not equal.

Example 7: Finding Limits by Direct Substitution:

lim (x + 4)/(x - 3)
x®1

Do the numerator separately from the denominator:

lim (x + 4) = 5
x®1

lim (x - 3) = -2 so
x®1

lim (x + 4)/(x - 3) = -5/2
x®1

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I. Conditions Under Which Limits Do not exist:

The Limit of f(x) as x ® c does not exist if any of the following conditions is true.

1. f(x) approaches a different number from the right side of "c" than from the left side of "c".

2. f(x) increases or decreases without bound as x approaches c.

3. f(x) oscillates between two fixed values as x approaches c.

II. Properties of Limits:

Let "b" and "c" be real numbers and let "n" be a positive integer.

1. lim b = b
x®c

2. lim x = c
x®c

3. lim xn = cn
x®c

4. lim n Ö(x) = nÖ(c) for n even and c greater than zero.
x®c

III. Operations with Limits:

Given: lim f(x) = L
x®c
and

lim g(x) = K
x®c

1. Scalar multiple:
lim [ bf(x)] = bL
x®c

2a. Sum of limits:
lim [ f(x) + g(x)] = L + K
x®c

2b. Difference of limits:
lim [ f(x) - g(x)] = L - K
x®c

3. Product:
lim [ f(x) ´ g(x)] = L ´ K
x®c

4. Quotient:
lim [ f(x) ¸ g(x)] = L ¸ K
x®c

5. Power:
lim [ f(x)]n = Ln
x®c

Example 8:

lim f(x) = 3x
x®c

and

lim g(x) = -2x + 1
x®c

Find:
lim [ f(x) + g(x)] =3(3) + (-2)(3) + 1 = 9 - 6 + 1 = 4
x®3

Find:
lim [ f(x) g(x)] = [(3)(4)][-2(4) + 1] = (12)(-8 + 1) = (12)(-7) = -84
x®4

Example 9:
lim f(x) = 3
x®c

and

lim g(x) = -2
x®c

Find:
lim [ f(x) + g(x)]2 = (3 + -2)2 = 1
x®c

Find:
lim [ 6f(x)g(x)] = 6(3)(-2) = -36
x®c